Question

Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil 12 pleces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in 0.0001 in.) is determined. The data appears in the table. Soil Type (B) 2 3 1 64 46 51 2 53 51 48 Coating (A)4544 51 41) 4 51 3 51 Complete the ANOVA table. Use α-0.05. (Round your answers to two decimal places.) Source dt what can you conclude for Fail to reject Hoa. The data does not suggest that there is a coating erect. O Fail to rejectHa. The data sugoests that there is a coating effect O Rejact a Tha data does not suggest that thera is a coating effact. O Reject Hoa The data suggests that there is a coating efTect. What can you conclude for Hop O Reject Hoa- The data does not suggest that there is a sail type effect O Reject Ho The data suggests that there is a soil type eftect. O Fail to reject Hog- The data does not suggest that there is a soil type efect OFail to reject Hur The data suggests that there is a soil type effect (b) Computa the model parametars. (Round your answars to two decimal places.) -2 a ,

Answer 1

using excel data analysis tool for one factor anova, following o/p Is obtained,

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
1	3	161	54	86.333
2	3	152	51	6.333
3	3	140	47	14.333
4	3	145	48	21.333
1	4	213	53	62.917
2	4	184	142	12.667
3	4	201	134	2.250
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Rows,A	83.00	3	27.67	1.10	0.42	4.76
Columns,B	106.17	2	53.08	2.12	0.20	5.14
Error	150.50	6	25.08
Total	339.67	11

p-value for A=0.42>0.05, so,
fail to reject Ho, the data does not suggests that there is a coating effect

p-value for B=0.20>0.05,so,
fail to reject Ho, the data does not suggests that there is a soil type effect

--------------------------------

b)

µ hat = Σx/n = 49.83

α hat = Xjbar - µhat

ßhat = Xi bar - µhat

α1 = 3.83

α2=0.84

α3=-3.16

α4=-1.50

ß1=3.42

ß2= -3.83

ß3=0.42